2017 ACM/ICPC Asia Regional Shenyang Online 1005 number number number(矩阵快速幂)

原题：

number number number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

We define a sequence F:

⋅ F0=0,F1=1;
⋅ Fn=Fn−1+Fn−2 (n≥2).

Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤ak, this positive number is mjf−good. Otherwise, this positive number is mjf−bad.
Now, give you an integer k, you task is to find the minimal positive mjf−bad number.
The answer may be too large. Please print the answer modulo 998244353.

 

Input

There are about 500 test cases, end up with EOF.
Each test case includes an integer k which is described above. (1≤k≤109)

 

Output

For each case, output the minimal mjf−bad number mod 998244353.

 

Sample Input

1

 

Sample Output

4

#include <iostream>#include <iomanip>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <deque>#include <stack>#include <string>#include <cmath>#include <vector>#include <utility>#include <set>#include <map>#include <sstream>#include <climits>//#pragma comment(linker, "/STACK:1024000000,1024000000")#define pi acos(-1.0)#define INF 2147483647using namespace std;typedef long long LL;typedef pair<int,int > P;const int MOD=998244353;struct Matrix{    LL a[100][100];    int r, c;} mat1,mat2;Matrix multi(Matrix x, Matrix y){    Matrix z;    memset(z.a, 0, sizeof(z.a));    z.r = x.r, z.c = y.c;    for(int i = 1; i <= x.r; i++)    {        for(int k = 1; k <= x.c; k++)        {            if(x.a[i][k] == 0)                continue;            for(int j = 1; j<= y.c; j++)                z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;        }    }    return z;}int Matrix_mod(int n){    while(n)    {        if(n & 1)            mat2 = multi(mat1, mat2);        mat1 = multi(mat1, mat1);        n >>= 1;    }    return mat2.a[1][2] % MOD;}int main(){    LL N;    while(scanf("%lld", &N)!=EOF)    {        memset(mat2.a, 0, sizeof(mat2.a));        mat2.r = 2, mat2.c = 2;        for(int i = 1; i <= 2; i++)            mat2.a[i][i] = 1;        mat1.r = 2,mat1.c = 2;        mat1.a[1][1] = mat1.a[1][2] = mat1.a[2][1] = 1;        mat1.a[2][2] = 0;        cout<<Matrix_mod(5+(N-1)*2)-1<<endl;    }    return 0;}